Implicit Differentiation

MATTechniques of Differentiation

When $y$ is not isolated but tied to $x$ through an equation, we differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ and applying the chain rule to every $y$ -term. Solving the resulting equation for $\dfrac{dy}{dx}$ gives the derivative.

ISC favours implicit equations involving variable exponents, symmetric relations, and trigonometric constraints requiring elegant proofs.

Implicit chain rule

\dfrac{d}{dx}\,[\,y^n\,] = n\,y^{\,n-1}\dfrac{dy}{dx}

Every term containing

y

picks up a factor

\dfrac{dy}{dx}

; collect these and solve.

Variable exponent case

x^{y} = y^{x} \;\Rightarrow\; y\log x = x\log y

Take logs first, then differentiate implicitly; gives

\dfrac{dy}{dx}=\dfrac{y(x\log y - y)}{x(y\log x - x)}

after rearranging.

Trigonometric implicit relation

\sin y = x\sin(a+y) \;\Rightarrow\; \dfrac{dy}{dx} = \dfrac{\sin^2(a+y)}{\sin a}

Write

x=\dfrac{\sin y}{\sin(a+y)}

, differentiate, and use

\sin(a+y)\cos y-\cos(a+y)\sin y=\sin a

Surd relation result

x\sqrt{1+y}+y\sqrt{1+x}=0 \;\Rightarrow\; \dfrac{dy}{dx} = -\dfrac{1}{(1+x)^2}

x\ne y

; obtained by reducing the relation to

y=-\dfrac{x}{1+x}

then differentiating.

Differentiate term by term, attaching $\dfrac{dy}{dx}$ to each $y$ -term via the chain rule.
Gather all $\dfrac{dy}{dx}$ terms on one side, factor it out, and divide to isolate the derivative.
For equations with $y$ in an exponent (e.g. $x^y=y^x$ ), take natural logarithms before differentiating.
Sometimes it is cleaner to solve the implicit relation for $y$ explicitly first (as in the surd problem) when feasible.
Use product and quotient rules on mixed terms such as $xy$ , which differentiates to $y+x\dfrac{dy}{dx}$ .
For proof-type questions, simplify using trig identities (e.g. angle-subtraction) rather than leaving the answer in $x$ .
The final $\dfrac{dy}{dx}$ may legitimately contain both $x$ and $y$ .

Where the marks go

Forgetting the $\dfrac{dy}{dx}$ factor when differentiating a $y$ -term, treating $y$ as a constant.
Differentiating $x^y$ as $y\,x^{y-1}$ (power rule) instead of taking logs for the variable exponent.
Mishandling the product rule on $xy$ or $x\sin(a+y)$ terms.
Leaving the answer un-simplified in a proof question when the target form (e.g. $\dfrac{\sin^2(a+y)}{\sin a}$ ) is required.

How the board asks it

Numericalimplicit chain rule on a symmetric relation
If $x^3 + y^3 = 3axy$ , find $\dfrac{dy}{dx}$ .
Numericallogarithmic differentiation, variable exponent
If $x^y = y^x$ , find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
Derive / provesurd relation, inverse-trig substitution
If $\sqrt{1-x^2} + \sqrt{1-y^2} = a(x-y)$ , show that $\dfrac{dy}{dx} = \sqrt{\dfrac{1-y^2}{1-x^2}}$ .
Numericallogarithmic differentiation of a product of variable exponents
If $x^y \cdot y^x = 1$ , find $\dfrac{dy}{dx}$ .
Numericaltrigonometric implicit relation
If $\sin y = x\sin(a+y)$ , find $\dfrac{dy}{dx}$ and evaluate it at $x = 0$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.